Chapter 13 Solutions Molarity.

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Presentation transcript:

Chapter 13 Solutions Molarity

Molarity (M) A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute 1 liter solution

Units of Molarity 2.0 M HCl = 2.0 moles HCl 1 L HCl solution 6.0 M HCl = 6.0 moles 1 L HCl solution

Math Problems Molarity is a conversion factor 3 types of M problems if you know the molarity 1 - You can determine the # of moles (or grams) solute in a given volume of solution. 2 - You can calculate the vol of solution that contains a specific # of moles solute if you know the molarity 3 If given the moles (grams) solute and vol of solution determine the molarity

Molarity Calculation NaOH is used to open stopped sinks, to treat cellulose in the making of nylon, and to remove potato peels commercially. If 4.0 g NaOH (mm = 40.0 g/mol) are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution? 1 - calc the moles solute Mol = g/mm 2 - Determine L of solution L = mL/1000 3 - M = mol solute/L solution

Calculating Molarity 1. 4.0 g NaOH x 1 mol NaOH = 0.10 mol NaOH 2. 500. mL x 1 L = 0.500 L 1000 mL 3. 0.10 mol NaOH = 0.20 mol NaOH 0.500 L 1 L = 0.20 M NaOH

Learning Check M1 A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution? 1) 8 M 2) 5 M 3) 2 M

Learning Check M1 A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution? 2) 5 M M = 2 mole KOH = 5 M 0.4 L

Learning Check A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1 - Determine moles of solute (glucose) 2 - Divide by L solution 1) 0.20 M 2) 5.0 M 3) 36 M

Solution A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 72 g x 1 mole = 0.40 mol = 0.20 M 180. g 2.0 L

Molarity Conversion Factors A solution is a 3.0 M NaOH. Write the molarity in the form of conversion factors that relates vol of solution to # of mol solute 3.0 moles NaOH and 1 L NaOH soln 1 L NaOH soln 3.0 moles NaOH

Learning Check Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution? 1 - determine L of solution 2 - convert from L to moles with molarity 1) 15 moles HCl 2) 1.5 moles HCl 3) 0.15 moles HCl

Solution 3) 1500 mL x 1 L = 1.5 L 1000 mL 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1 L (Molarity is a conversion factor)

Learning Check How many grams of KCl are present in 2.5 L of 0.50 M KCl? 1 - convert from L to moles solute, KCl 2 - convert moles to grams of KCl 1) 1.3 g 2) 5.0 g 3) 93 g

Solution 3) 2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl 1 L 1 mole KCl

Learning Check M5 How many liters of stomach acid, which is 0.10 M HCl, contain 0.15 mole HCl? 1 - Convert from mol to L with molarity 1) 0.150 L 2) 1.500 L 3) 5.000 L

Solution M5 0.15 mol HCl x 1 L soln 0.10 mol HCl (Molarity inverted) 2) 0.15 mol HCl x 1 L soln 0.10 mol HCl (Molarity inverted) = 1.500 L HCl

Learning Check M6 How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 1 - Deter L of solution 2 - Convert L to moles with molarity 3 - Convert mol to grams with molar mass 1) 12 g 2) 48 g 3) 300 g

Solution 2) 400. mL x 1 L = 0.400 L 1000 mL 0.400 L x 3.0 mol NaOH x 40.0 g NaOH 1 L 1 mol NaOH (molarity) (molar mass) = 48 g NaOH